MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 25
Last lecture we introduced the notion of a principal ideal in a commutative ring. These are ideals in a commutative ring R that take the form
paq “ aR “ tar ∶ r P Ru
A principal ideal domain (PID) is an integral domain R (a commutative ring such that ab “ 0 implies a “ 0 or b “ 0) such that every ideal in R is principal.
The ring of integers Z is the most basic example of a PID. The ideals in Z are p0q, p1q, p2q, p3q, p4q,...
PIDS are an important class of rings because they occur frequently and the theory of PIDS is considerably simpler than that of general (commutative) rings.
§ Let R be a non-zero commutative ring. The following are equivalent: (i) R is a field.
(ii) The only ideals in R are p0q “ t0u and p1q “ R.
(iii) Every homomorphism φ ∶ R Ñ R1 where R1 ‰ t0u is 1-1.
Proof. We show (i) implies (ii). Assume (i), i.e. R is a division ring. Consider an ideal I Ă R with I ‰ p0q. Choose a P I, a ‰ 0. Since R is a division ring, a´1 P R. Then 1 “ a´1a P I as a´1 P R and a P I. Now for any b P R, we have b “ b1 P I as b P R and 1 P I. Thus (ii) holds.
Next, we show (ii) implies (iii). Assume (ii): the only ideals in R are t0u and R. Consider a homomorphism φ ∶ R Ñ R1 where R1 ‰ t0u. Then kerpφq Ă R is a proper ideal as 1 R kerpφq. By our assumption it must be t0u. This is equivalent to φ being 1-1. Thus (iii) holds.
Finally, we show (iii) implies (i). Assume (iii), i.e. every homomorphism φ ∶ R Ñ R1 where R1 ‰ t0u is 1-1. Now take a P R with a ‰ 0. Consider the natural homomorphism φ ∶ R Ñ R{paq. Suppose R{paq ‰ t0u, so that φ is 1-1. Then kerpφq “ p0q. On the other hand, kerpφq “ paq. Then paq “ p0q implies a “ 0, a contradiction. Thus R{paq “ t0u, impying paq “ R “ p1q. In particular, 1 “ ar for some r P R, so a has a multiplicative inverse. We have shown that every non-zero a P R is invertible, and thus R is a field.
A corollary of this result is the following:
§ If R is a field, then R is a PID.
This holds simply because the only ideals in a field R are the ideals p0q “ t0u and p1q “ R which are principal ideals.
1 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 25
Thus the fields Q, R, C are all examples of PIDs.
A more exotic example of a PID is the ring of Gaussian integers
Zris “ ta ` bi ∶ a, b P Zu Ă C We will show this is a PID later. ? We saw last lecture? that the ring Zr ´3s is not a PID, because we showed that the ideal consisting of a ` b ´3 with a ” b (mod 2) is not principal.
Polynomial rings Another important example involves the following construction. Let R be any ring. Define
Rrxs “ tpolynomials in x with coefficients in Ru
That is, a typical element in Rrxs is an expression of the form
n n´1 fpxq “ anx ` an´1x ` ⋯a1x ` a0 where n is a non-negative integer and a0, . . . , an are elements of the ring R. The sum fpxq ` gpxq and product fpxqgpxq of two such polynomials are defined in the usual fashion, and this makes Rrxs into a ring. If R is commutative, so is Rrxs. Also, if R is an integral domain, so too is the polynomial ring Rrxs.
§ If R is a field, then Rrxs is a PID.
Before explaining the proof, let’s explore the consequences of this result.
The rings Qrxs, Rrxs, Crxs are all PIDS. This means, for example, that for any ideal I Ă Qrxs there exists a polynomial fpxq with rational coefficients such that I “ pfpxqq. The polyno- mial fpxq is unique up to multiplication by an element in Qˆ.
Let p be a prime. Then Zp is a field, so Zprxs is a PID. For example, consider Z2rxs, polynomials with coefficients in Z2. Some elements in this ring are 0, 1, x, 1 ` x, x2, 1 ` x2, 1 ` x ` x2,...
For any ideal I Ă Z2rxs we can find some such element fpxq such that I “ pfpxqq.
The condition that R be a field in the result is necessary. For example, consider Zrxs. Of course Z is not a field, so the result does not apply here. In fact Zrxs is not a PID: the ideal
I “ t2fpxq ` xgpxq ∶ fpxq, gpxq P Zrxsu Ă Zrxs is not a principal ideal, as you can verify.
2 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 25
Now let us prove the statement. We will need the following:
§ (Division algorithm for polynomials) Let R be a field. Consider polynomials fpxq and gpxq ‰ 0 in Rrxs. Then there exist unique qpxq, rpxq P Rrxs such that
fpxq “ gpxqqpxq ` rpxq and where either degprpxqq ă degpgpxqq or rpxq “ 0.
i Here the degree of fpxq “ aix P Rrxs is the largest n such that an ‰ 0. We omit the proof of the division algorithm and now show that if R is a field then Rrxs is a PID. ř Let I Ă Rrxs be a PID. We must show I is principal. If I “ t0u then it is the principal ideal p0q, so assume I ‰ p0q. Let gpxq P I be non-zero and of minimal possible degree among all non-zero polynomials in I. Note pgpxqq Ă I. Now consider any other element fpxq P I. Then the division algorithm gives us qpxq, rpxq P Rrxs satisyfing
fpxq “ gpxqqpxq ` rpxq and degprpxqq ă degpgpxqq or rpxq “ 0. Suppose rpxq ‰ 0. Then
rpxq “ fpxq ´ gpxqqpxq is in I, because fpxq, gpxq P I and qpxq P Rrxs. Furthermore, degprpxqq ă degpgpxqq. But this contradicts our assumption that gpxq has the minimal possible degree among non-zero polynomials in I. So we must have rpxq “ 0. Then
fpxq “ qpxqqpxq
This shows fpxq P pgpxqq. Thus I “ pgpxqq, and I is a principal ideal. This completes the proof that all ideals in Rrxs are principal.
Finally, we remark that if you continue to add variables to your ring, and consider polyno- mials in several variables, you will not get a PID. For example, consider
Qrxsrys “ Qrx, ys “ tpolynomials in x, y with coefficients in Qu
Then the ideal generated by the polynomials x and y, which is given by I “ txfpx, yq ` ygpx, yq ∶ fpx, yq, gpx, yq P Qrxsu, is not a principal ideal in Qrx, ys.
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